Devoweling the vowelful using user input. What the heck is wrong with my code?

Question

I am attempting to write my own version of the devowel.py program that Kenneth wrote. I did a lot of Googling and research on Stack Overflow to get where I currently am. The program runs without kicking errors, but unfortunately my while loop runs infinitely and it does not remove the vowels.

When I run it piece by piece in the interpreter my bool methods work. However, since my loop isn't ending, I'm assuming it's poor looping on my part. Does anyone see what may be the cause?

>>> input_word = input("Please enter any word, and we'll get rid of those pesky vowels: ")
Please enter any word, and we'll get rid of those pesky vowels: trhs
>>> vowels_bool = bool(vowels.intersection(input_word))
>>> vowels_bool                                                                 
False
>>> input_word = input("Please enter any word, and we'll get rid of those pesky vowels: ")
Please enter any word, and we'll get rid of those pesky vowels: treehouse
>>> vowels_bool = bool(vowels.intersection(input_word))                         
>>> vowels_bool                                                                 
True

In addition to the looping error. My remove vowels set does not work. I did some digging and it looks like it should work, but it's not.

I appreciate all constructive criticism!

Disclaimer: I know the code is ugly.

# I want the user to enter a word
input_word = input("Please enter any word, and we'll get rid of those pesky vowels: ")
# I want to tell the computer what a vowel is
vowels = set(['a', 'e', 'i', 'o', 'u', 'y'])
# I want to find out if my word has vowels
vowels_bool = bool(vowels.intersection(input_word))
# I need to convert user input to list form
input_word_list = list(input_word)
# I want to remove the vowels
vowels_removed = [v for i, v in enumerate(input_word_list) if i not in vowels]
# I want to print the word without the vowels and the first word capitalized
vowels_removed_string = ''.join(vowels_removed).capitalize()

while vowels_bool == True:
    try:
        vowels_removed = [v for i, v in enumerate(input_word_list) if i not in vowels]
    except: 
        if vowels_bool == False:
            print("Here's your vowelless word: {}".format(vowels_removed_string))
            break

Answer 1 · 2015-04-19T09:27:34Z

April 19, 2015 9:27am

Your vowels_bool variable never gets updated in the loop so it will always remain True. In your list comprehensions you're checking to see if the index/count is in vowels instead of a letter, as enumerate generates a list of tuples in the form (count, value).

Using list comprehensions you can do the following:

VOWELS = ['a', 'e', 'i', 'o', 'u']

response = input("Enter a word: ")
result = ''.join([letter for letter in response if letter not in VOWELS])

print(result)

Or a similar way that I find easier to read:

VOWELS = ['a', 'e', 'i', 'o', 'u']

response = input("Enter a word: ")
result = filter(lambda letter: letter not in VOWELS, response)

print(result)

Answer 2 · 2015-04-19T09:14:24Z

April 19, 2015 9:14am

Since you only need to check each letter in the word once, I recommend using 'for .. ' instead of a while loop. Here's how I would do it:

# I want the user to enter a word and instantly turn it into a list
input_word = list(raw_input('What is your name? '))

# I want to tell the computer what a vowel is
vowels = ['A', 'E', 'I', 'O', 'U', 'Y']

#function to loop over each letter in the word. If it is in the vowel list, remove it.
def remover(word, vowel_list):
    for letter in word:
        if letter.upper() in vowel_list:
            word.remove(letter)
    return ''.join(word).capitalize()

#run the function
devoweled = remover(input_word, vowels)

#print("Here's your vowelless word: {}".format())
print("Here's your vowelless word: {}".format(devoweled))

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Aaron Brown

Aaron Brown

Devoweling the vowelful using user input. What the heck is wrong with my code?

2 Answers

Dan Johnson

Dan Johnson

Aaron Brown

Aaron Brown

Alan Dsouza

Alan Dsouza

Aaron Brown

Aaron Brown